MATH SOLVE

5 months ago

Q:
# Cydney is proving that perpendicular lines have slopes that are opposite reciprocals. She draws line p and labels two points on the line as (0, a)(0, a) and (b, 0)(b, 0) . Enter the answers, in simplest form, in the boxes to complete the proof. The slope of line p is . Rotate line p 90° counterclockwise about the origin to get line q. The labeled points on line p map to (−a, 0)(−a, 0) and ( , ) on line q. The slope of line q is . The slopes of the lines are opposite reciprocals because the product of the slopes is .

Accepted Solution

A:

check the picture below.

with a rotation of 90° from the origin, we get from points (0, a) and (b,0) to the points with equivalent values of (-a, 0) and (0, b). Notice that in the graph we can map those values to a = 4 and b = 6, however, "a" and "b" are just constants and can be any value whatsoever, in this case for the sake of exemplification, they happen to be that.

now, slopes of perpendicular lines, have a product of -1, let's check these ones,

[tex]\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~{{ 0}} &,&{{ a}}~) % (c,d) &&(~{{ b}} &,&{{ 0}}~) \end{array} \\\\\\ % slope = m slope = {{ m}}\implies \cfrac{\stackrel{rise}{{{ y_2}}-{{ y_1}}}}{\stackrel{run}{{{ x_2}}-{{ x_1}}}}\implies \cfrac{0-a}{b-0}\implies \cfrac{-a}{b}\\\\ -------------------------------\\\\[/tex]

[tex]\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~{{ -a}} &,&{{ 0}}~) % (c,d) &&(~{{ 0}} &,&{{ b}}~) \end{array} \\\\\\ % slope = m slope = {{ m}}\implies \cfrac{\stackrel{rise}{{{ y_2}}-{{ y_1}}}}{\stackrel{run}{{{ x_2}}-{{ x_1}}}}\implies \cfrac{b-0}{0-(-a)}\implies \cfrac{b-0}{0+a}\implies \cfrac{b}{a}\\\\ -------------------------------\\\\ \textit{their product will then be }\quad \cfrac{-a}{b}\cdot \cfrac{b}{a}\implies -1[/tex]

now, we could check using the a = 4 and b = 6 values as well,

[tex]\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~{{ 0}} &,&{{ 4}}~) % (c,d) &&(~{{ 6}} &,&{{ 0}}~) \end{array} \\\\\\ % slope = m slope = {{ m}}\implies \cfrac{\stackrel{rise}{{{ y_2}}-{{ y_1}}}}{\stackrel{run}{{{ x_2}}-{{ x_1}}}}\implies \cfrac{0-4}{6-0}\implies \cfrac{-4}{6}\implies \cfrac{-2}{3}\\\\ -------------------------------\\\\[/tex]

[tex]\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~{{ -4}} &,&{{ 0}}~) % (c,d) &&(~{{ 0}} &,&{{ 6}}~) \end{array} \\\\\\ % slope = m slope = {{ m}}\implies \cfrac{\stackrel{rise}{{{ y_2}}-{{ y_1}}}}{\stackrel{run}{{{ x_2}}-{{ x_1}}}}\implies \cfrac{6-0}{0-(-4)}\implies \cfrac{6}{0+4}\implies \cfrac{3}{2}\\\\ -------------------------------\\\\ \textit{their product will then be }\quad \cfrac{-2}{3}\implies \cfrac{3}{2}\implies -1[/tex]

with a rotation of 90° from the origin, we get from points (0, a) and (b,0) to the points with equivalent values of (-a, 0) and (0, b). Notice that in the graph we can map those values to a = 4 and b = 6, however, "a" and "b" are just constants and can be any value whatsoever, in this case for the sake of exemplification, they happen to be that.

now, slopes of perpendicular lines, have a product of -1, let's check these ones,

[tex]\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~{{ 0}} &,&{{ a}}~) % (c,d) &&(~{{ b}} &,&{{ 0}}~) \end{array} \\\\\\ % slope = m slope = {{ m}}\implies \cfrac{\stackrel{rise}{{{ y_2}}-{{ y_1}}}}{\stackrel{run}{{{ x_2}}-{{ x_1}}}}\implies \cfrac{0-a}{b-0}\implies \cfrac{-a}{b}\\\\ -------------------------------\\\\[/tex]

[tex]\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~{{ -a}} &,&{{ 0}}~) % (c,d) &&(~{{ 0}} &,&{{ b}}~) \end{array} \\\\\\ % slope = m slope = {{ m}}\implies \cfrac{\stackrel{rise}{{{ y_2}}-{{ y_1}}}}{\stackrel{run}{{{ x_2}}-{{ x_1}}}}\implies \cfrac{b-0}{0-(-a)}\implies \cfrac{b-0}{0+a}\implies \cfrac{b}{a}\\\\ -------------------------------\\\\ \textit{their product will then be }\quad \cfrac{-a}{b}\cdot \cfrac{b}{a}\implies -1[/tex]

now, we could check using the a = 4 and b = 6 values as well,

[tex]\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~{{ 0}} &,&{{ 4}}~) % (c,d) &&(~{{ 6}} &,&{{ 0}}~) \end{array} \\\\\\ % slope = m slope = {{ m}}\implies \cfrac{\stackrel{rise}{{{ y_2}}-{{ y_1}}}}{\stackrel{run}{{{ x_2}}-{{ x_1}}}}\implies \cfrac{0-4}{6-0}\implies \cfrac{-4}{6}\implies \cfrac{-2}{3}\\\\ -------------------------------\\\\[/tex]

[tex]\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~{{ -4}} &,&{{ 0}}~) % (c,d) &&(~{{ 0}} &,&{{ 6}}~) \end{array} \\\\\\ % slope = m slope = {{ m}}\implies \cfrac{\stackrel{rise}{{{ y_2}}-{{ y_1}}}}{\stackrel{run}{{{ x_2}}-{{ x_1}}}}\implies \cfrac{6-0}{0-(-4)}\implies \cfrac{6}{0+4}\implies \cfrac{3}{2}\\\\ -------------------------------\\\\ \textit{their product will then be }\quad \cfrac{-2}{3}\implies \cfrac{3}{2}\implies -1[/tex]